Continue statement in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of loop with continue

What is the output of the following C++ code?

C++

#include <iostream>
int main() {
    for (int i = 1; i <= 5; i++) {
        if (i == 3) continue;
        std::cout << i << " ";
    }
    return 0;
}

A 1 2 4 5

B 2 3 4 5

C 1 2 3 4

D 1 2 3 4 5

💡 Hint

The continue statement skips the rest of the loop body for the current iteration.

When i equals 3, the continue statement skips printing 3 and moves to the next iteration. So the output excludes 3.

❓ Predict Output

intermediate

Sum with continue skipping multiples of 4

What is the value of sum after running this code?

C++

#include <iostream>
int main() {
    int sum = 0;
    for (int i = 1; i <= 10; i++) {
        if (i % 4 == 0) continue;
        sum += i;
    }
    std::cout << sum;
    return 0;
}

A 43

B 50

C 40

D 55

💡 Hint

Numbers divisible by 4 are skipped and not added to sum.

Numbers 4 and 8 are skipped. Sum of 1+2+3+5+6+7+9+10 = 43.

❓ Predict Output

advanced

Nested loops with continue

What is the output of this nested loop code?

C++

#include <iostream>
int main() {
    for (int i = 1; i <= 3; i++) {
        for (int j = 1; j <= 3; j++) {
            if (j == 2) continue;
            std::cout << i << j << " ";
        }
    }
    return 0;
}

A 12 22 32

B 11 12 13 21 22 23 31 32 33

C 11 13 22 23 31 33

D 11 13 21 23 31 33

💡 Hint

The continue skips printing when j equals 2 inside inner loop.

For each i, j=2 is skipped, so only j=1 and j=3 print. Output is pairs: 11 13 21 23 31 33.

❓ Predict Output

advanced

Continue in while loop

What is the output of this code?

C++

#include <iostream>
int main() {
    int i = 0;
    while (i < 5) {
        i++;
        if (i == 3) continue;
        std::cout << i << " ";
    }
    return 0;
}

A 1 2 3 4 5

B 1 2 4 5

C 2 3 4 5

D 1 2 3 4

💡 Hint

The continue skips printing when i equals 3 after increment.

When i is 3, continue skips printing 3. So output excludes 3.

🧠 Conceptual

expert

Effect of continue on loop variable increment

Consider this code snippet. What will be the output?

C++

#include <iostream>
int main() {
    for (int i = 0; i < 5; ) {
        i++;
        if (i == 3) continue;
        std::cout << i << " ";
    }
    return 0;
}

A 1 2 3 4 5

B 1 2 3 4

C 1 2 4 5

D 2 3 4 5

💡 Hint

The increment is inside the loop body before continue.

i is incremented before continue, so when i == 3, continue skips printing 3. Output excludes 3.

Practice

1. What does the continue statement do inside a loop in C++?

easy

A. It stops the entire loop immediately.

B. It skips the rest of the current loop iteration and moves to the next iteration.

C. It exits the current function.

D. It restarts the program from the beginning.

Solution

1. Step 1: Understand the role of continue in loops

   The continue statement tells the program to skip the remaining code in the current loop cycle.

2. Step 2: Identify what happens after continue

   After skipping, the loop moves to the next iteration immediately without executing the rest of the loop body.

3. Final Answer:

   It skips the rest of the current loop iteration and moves to the next iteration. -> Option B

4. Quick Check:

   Continue skips current iteration [OK]

Hint: Continue skips current loop step, not whole loop [OK]

Common Mistakes:

• Thinking continue stops the loop completely
• Confusing continue with break
• Assuming continue exits the function

2. Which of the following is the correct syntax to use continue inside a for loop in C++?

easy

A. continue;

B. continue()

C. continue{}

D. continue[]

Solution

1. Step 1: Recall the syntax of continue in C++

   The continue statement is a keyword and used alone followed by a semicolon.

2. Step 2: Check each option's syntax

   Only continue; is valid. The others use parentheses or braces which are incorrect.

3. Final Answer:

   continue; -> Option A

4. Quick Check:

   Continue ends with semicolon [OK]

Hint: Continue is a keyword, ends with semicolon [OK]

Common Mistakes:

• Adding parentheses after continue
• Using braces or brackets with continue
• Omitting the semicolon

3. What will be the output of the following C++ code?

for (int i = 1; i <= 5; i++) {
    if (i == 3) continue;
    std::cout << i << " ";
}

medium

A. 1 2 3 4 5

B. 1 2 3 4

C. 1 2 4 5

D. 2 3 4 5

Solution

1. Step 1: Analyze the loop and condition

   The loop runs from 1 to 5. When i == 3, the continue skips printing.

2. Step 2: Determine printed values

   Numbers 1 and 2 print normally. At 3, continue skips printing. Then 4 and 5 print.

3. Final Answer:

   1 2 4 5 -> Option C

4. Quick Check:

   Skip 3, print others [OK]

Hint: Continue skips printing 3, others print [OK]

Common Mistakes:

• Including 3 in output
• Stopping loop at 3
• Printing numbers after 3 incorrectly

4. Identify the error in the following C++ code snippet:

for (int i = 0; i < 5; i++) {
    if (i == 2)
        continue
    std::cout << i << " ";
}

medium

A. No error in the code.

B. Incorrect loop condition.

C. Wrong variable type for i.

D. Missing semicolon after continue statement.

Solution

1. Step 1: Check syntax of continue statement

   The continue statement must end with a semicolon.

2. Step 2: Identify missing semicolon

   In the code, continue is not followed by a semicolon, causing a syntax error.

3. Final Answer:

   Missing semicolon after continue statement. -> Option D

4. Quick Check:

   Continue needs semicolon [OK]

Hint: Always put semicolon after continue [OK]

Common Mistakes:

• Forgetting semicolon after continue
• Assuming continue needs parentheses
• Misreading loop condition as error

5. Consider the following code that sums numbers from 1 to 10 but skips multiples of 3 using continue. What is the final sum?

int sum = 0;
for (int i = 1; i <= 10; i++) {
    if (i % 3 == 0) continue;
    sum += i;
}
std::cout << sum;

hard

A. 52

B. 55

C. 45

D. 50

Solution

1. Step 1: Identify numbers skipped by continue

   Numbers divisible by 3 (3, 6, 9) are skipped and not added to sum.

2. Step 2: Calculate sum of numbers 1 to 10 excluding 3, 6, 9

   Sum 1+2+4+5+7+8+10 = 37 + 15 = 52

3. Final Answer:

   52 -> Option A

4. Quick Check:

   Sum excluding multiples of 3 = 52 [OK]

Hint: Skip multiples of 3, sum others [OK]

Common Mistakes:

• Including multiples of 3 in sum
• Adding all numbers 1 to 10
• Misunderstanding continue effect